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Numbers $1,2,3, \ldots ,2n(n \in N)$are printed on $2n$ cards. The probability of drawing a number $n$ is proportional to $r$. Then find the probability of drawing an even number in one draw.
A.$\dfrac{{n + 2}}{{n + 3}}$
B.$\dfrac{{n + 1}}{{n + 3}}$
C.$\dfrac{1}{2}$
D.$\dfrac{{n + 1}}{{2n + 1}}$

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Last updated date: 14th Jun 2024
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Answer
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Hint: According to the problem there are $2n$ cards. The probability of drawing a number is proportional to $r$. The probability P(r) that the number $r$ is drawn in one draw is given by $P(r) = kr$.
Now, we find the probability P(r) for each number $1,2,3, \ldots ,2n(n \in N)$ is drawn in one draw by substituting $r = 1,2,3, \ldots ,2n(n \in N)$ into $P(r) = kr$.
From the postulate of probability that is the sum of the probabilities of all outcomes must equal 1, we find the value of k.
$ \Rightarrow P(1) + P(2) + P(3) + \ldots + P(2n) = 1$
Use the formula to find the sum of n natural number that is,
$1 + 2 + 3 + \ldots + n = \dfrac{{n(n + 1)}}{2}$
The sum of 2n natural number is obtained by substituting 2n into the formula of the sum of n natural number,
$1 + 2 + 3 + \ldots + 2n = n(2n + 1)$
In the given question we are asked to find the probability of drawing an even number in one draw.
we have to find the sum of probability P(r) for even numbers is drawn in one draw.
$ \Rightarrow P(2) + P(4) + P(6) + \ldots + P(2n)$
Substitute the value of k and sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$.

Complete step-by-step solution:
There are $2n$ cards, at each card the numbers printed as $1,2,3, \ldots ,2n(n \in N)$.
The probability of drawing a number is directly proportional to the $r$. That means is the probability P(r) that the number $r$ is drawn in one draw.
$ \Rightarrow P(r) = kr$
Where k is the proportionality constant.
Determine the probability for the values of $r$ as $1,2,3, \ldots ,2n(n \in N)$.
The probability of getting one is printed on the card is,
$P(1) = k(1)$
The probability of getting two is printed on the card is,
$P(2) = 2k$
The probability of getting three is printed on the card is,
$P(3) = 3k$
Similarly, the probability of getting $2n$ is printed on the card is,
$P(2n) = 2nk$
The sum of the probabilities of all outcomes must equal 1. $\therefore $The addition of the probability of getting $1,2,3, \ldots ,2n(n \in N)$ is equal to 1.
$ \Rightarrow P(1) + P(2) + P(3) + \ldots + P(2n) = 1$
$ \Rightarrow k + 2k + 3k + \ldots + 2nk = 1$
Take common term k outside from the left side of the equation,
$ \Rightarrow k(1 + 2 + 3 + \ldots + 2n) = 1 \ldots (1)$
Now, we can find the sum of 2n natural numbers.
Since the sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$.
To find sum of 2n natural numbers that is,$1 + 2 + 3 + \ldots + 2n$ is found by substituting n as 2n into the formula $\dfrac{{n(n + 1)}}{2}$.
$1 + 2 + 3 + \ldots + 2n = \dfrac{{2n(2n + 1)}}{2}$
$1 + 2 + 3 + \ldots + 2n = n(2n + 1) \ldots (2)$
Substitute the value of $n(2n + 1)$ into the equation (1),
$ \Rightarrow kn(2n + 1) = 1$
Solve the equation and find the value of k,
$ \Rightarrow k = \dfrac{1}{{n(2n + 1)}}$
According to the given question we are asked to find the probability of drawing an even number in one draw. So determine the addition of the probability for the even numbers,$ \Rightarrow P(2) + P(4) + P(6) + \ldots + P(2n)$
Since $P(r) = kr$ therefore,
$P(2) = 2k$
$P(4) = 4k$
$P(6) = 6k$
$ \Rightarrow P(2) + P(4) + P(6) + \ldots + P(2n)$
$ \Rightarrow 2k + 4k + 6k + \ldots + 2nk$
Take common term 2k outside from the left side of the equation,
$ \Rightarrow 2k(1 + 2 + 3 + \ldots + n)$
The sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$.
$ \Rightarrow 2k \times \dfrac{{n(n + 1)}}{2}$
Substitute $k = \dfrac{1}{{n(2n + 1)}}$ to get the probability of drawing an even number in one draw.
$ \Rightarrow 2 \times \dfrac{1}{{n(2n + 1)}} \times \dfrac{{n(n + 1)}}{2}$
Simplify the calculation,
$ \Rightarrow \dfrac{{n + 1}}{{2n + 1}}$

Option D is the correct answer.

Note: The most important step is to find the proportionality constant k. This can be found by the postulate, the sum of the probabilities of all outcomes must equal 1.
The formula that has to remember the sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$. By substituting 2n into the formula we get the sum of 2n natural numbers, $1 + 2 + 3 + \ldots + 2n$ is $\dfrac{{2n(2n + 1)}}{2}$ or $n(n + 1)$.